 Two-chain noncontinuous complete partial order

Take two increasing countable chains sharing a least element $\perp$ and a greatest element $\top$, with interiors $a_1<a_2<\cdots$ and $b_1<b_2<\cdots$ and no comparisons between branches. Each entire interior chain has $\top$ as its least upper bound. A directed set meeting both interiors must contain $\top$. Otherwise it lies within one branch, where it has either a greatest member or least upper bound $\top$. This proves that the partially ordered set is a directed-complete partial order.

The only element way-below $a_n$ is $\perp$. Indeed, an element way-below $a_n$ must be below $a_n$, but every positive $a_m\le a_n$ fails the way-below relation when tested against $\{b_1,b_2,\dots \}$: its least upper bound is $\top \ge a_n$, yet it has no member above $a_m$. Nonempty directed sets always contain an element above $\perp$, so $\perp \ll a_n$. Thus the elements way-below $a_n$ have least upper bound $\perp \ne a_n$, and this is not a continuous poset.